已知实数x满足x-x分之1=根号5.且x<0,求x+x分之1的值已知实数a、b满足a+b=8,ab=15,且a>b,求a-b的值 ∵a+b=8,ab=15 ∴(a-b)^2=a^2-2ab+b^2 =a^2+2ab+b^2-4ab =(a+b)^2-4ab =8^2-4×15=4 ∵a>b,∴a-b>0,∴a-b=2.
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![已知实数x满足x-x分之1=根号5.且x<0,求x+x分之1的值已知实数a、b满足a+b=8,ab=15,且a>b,求a-b的值 ∵a+b=8,ab=15 ∴(a-b)^2=a^2-2ab+b^2 =a^2+2ab+b^2-4ab =(a+b)^2-4ab =8^2-4×15=4 ∵a>b,∴a-b>0,∴a-b=2.](/uploads/image/z/2002349-29-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0x%E6%BB%A1%E8%B6%B3x-x%E5%88%86%E4%B9%8B1%3D%E6%A0%B9%E5%8F%B75.%E4%B8%94x%EF%BC%9C0%2C%E6%B1%82x%2Bx%E5%88%86%E4%B9%8B1%E7%9A%84%E5%80%BC%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0a%E3%80%81b%E6%BB%A1%E8%B6%B3a%2Bb%3D8%2Cab%3D15%2C%E4%B8%94a%3Eb%EF%BC%8C%E6%B1%82a-b%E7%9A%84%E5%80%BC+%E2%88%B5a%2Bb%3D8%2Cab%3D15+%E2%88%B4%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2+%3Da%5E2%2B2ab%2Bb%5E2-4ab+%3D%28a%2Bb%29%5E2-4ab+%3D8%5E2-4%C3%9715%3D4+%E2%88%B5a%3Eb%2C%E2%88%B4a-b%3E0%2C%E2%88%B4a-b%3D2.)
已知实数x满足x-x分之1=根号5.且x<0,求x+x分之1的值已知实数a、b满足a+b=8,ab=15,且a>b,求a-b的值 ∵a+b=8,ab=15 ∴(a-b)^2=a^2-2ab+b^2 =a^2+2ab+b^2-4ab =(a+b)^2-4ab =8^2-4×15=4 ∵a>b,∴a-b>0,∴a-b=2.
已知实数x满足x-x分之1=根号5.且x<0,求x+x分之1的值
已知实数a、b满足a+b=8,ab=15,且a>b,求a-b的值 ∵a+b=8,ab=15 ∴(a-b)^2=a^2-2ab+b^2 =a^2+2ab+b^2-4ab =(a+b)^2-4ab =8^2-4×15=4 ∵a>b,∴a-b>0,∴a-b=2.
已知实数x满足x-x分之1=根号5.且x<0,求x+x分之1的值已知实数a、b满足a+b=8,ab=15,且a>b,求a-b的值 ∵a+b=8,ab=15 ∴(a-b)^2=a^2-2ab+b^2 =a^2+2ab+b^2-4ab =(a+b)^2-4ab =8^2-4×15=4 ∵a>b,∴a-b>0,∴a-b=2.
x-1/x=根号5,
两边平方:x^2-2+1/x^2=5,
于是(两边加4)x^2+2+1/x^2=9,即(x+1/x)^2=9.
因为x<0,于是x+1/x<0,
因此上式两边开方,x+1/x取负值,x+1/x=-3.
(a-b)²
=a²+b²-2ab
=(a+b)²-4ab a+b=8 ab=15 代入
=64-60
=4
则a-b= ± 2
因为原题中a>b 则a-b>0
a-b= 2
x+1/x=8 x* 1/x=1
(x-1/x)²
=(x+1/x)²- 4*x* 1/x
=8-4=4
x-1/x =±2